Pushing the Parabola Robert I. Knighton Miami High School January 07, 2013

Consider a parabola in the x, y plane with the equation y = ax2 . Let us write this parabola in parametric form as follows: x=t y = at2 If this parabola were the perimeter of a physical object, and the region the x axis were a surface on which the object lied, and the parabola was given a ”push” towards the positive x, what would be the equation for the curve traced out by the vertex of the parabola, given that it rolled on forever? In visualizing the situation, it is clear that the x value of the point of contact between the parabola and the x axis is the arc length of the parabola from the vertex at that point. So, we are naturally led to define a function s(t), which is the arclength along the path of the parabola from the the origin to some (x(t), y(t)). It follows that Z tp s(t) = 1 + 4a2 x2 dx 0

We can make a substitution in order to simplify this a bit. Let 2ax = sinh θ So that dx =

cosh θ dθ 2a

So we can now write Z x=t p Z x=t 1 1 2 s(t) = 1 + (sinh θ) cosh θdθ = (cosh θ)2 dθ 2a x=0 2a x=0 1

(1)

1 = 4a

Z

x=t

1 + cosh 2θdθ = [ x=0

sinh2θ x=t θ + ] 4a 8a x=0

From (1), we can rewrite this as √ arsinh(2at) + 2at 1 + 4a2 t2 s(t) = 4a √ And since it can be shown that arsinh(v) = ln(v + 1 + v 2 ), we have √ √ ln(2at + 1 + 4a2 t2 ) t 1 + 4a2 t2 s(t) = + 4a 2

(2)

This is something we will return to later. Now if the parabola is standing up, visualizing the situation again, it is clear that if we define an angle φ to be the angle between the tangent line to the parabola at P = (t, at2 ), and the secant line from (0, 0) to P on the parabola, then if the parabola is later tilted such that the x axis is tangent to P , then the vertex will lie at the point x(t) = s(t) − R cos (φ(t))

(3)

y(t) = R sin (φ(t))

(4)

Where R is the distance √ between P and √ the vertex. 2 2 4 It is clear that R = t + a t = |t| 1 + a2 t2 , and s(t) is already known. So, now all that is needed is an expression φ(t). If we construct two vectors, one in the direction of the tangent line mentioned above, and one in the direction of the secant line, and dot the two, it is possible to find the angle between them by the theorem a · b = |ab| cos φ. It is quite simple to derive the tangent and secant√ vectors to be < 1, 2at > √ and < t, at2 > with magnitudes 1 + 4a2 t2 and t 1 + a2 t2 , respectively. The dot product of the two vectors is also trivial to find, and can be expressed as t + 2a2 t3 . Now, using the dot product theorem, we can solve for φ. p p t 1 + a2 t2 1 + 4a2 t2 cos φ = t + 2a2 t3 1 + 2a2 t2 cos φ = p (1 + a2 t2 )(1 + 4a2 t2 ) φ = arccos p

1 + 2a2 t2 (1 + a2 t2 )(1 + 4a2 t2 ) 2

(5)

Substituting (5) and (2) into (3) and (4), we get (albeit after a lot of simplification) √ √ ln(2at + 1 + 4a2 t2 ) t 1 + 4a2 t2 t(1 + 2a2 t2 ) x(t) = (6) + − √ 4a 2 1 + 4a2 t2 r (1 + 2a2 t2 )2 y(t) = |t| 1 + a2 t2 − (7) 1 + 4a2 t2 Which successfully traces the path that the vertex would take, if the parabola was given a push (to either side).

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